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1 year ago

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23.It is possible to dilute a solution by...Select one:a. adding more solute to the solution.b. adding more solvent to the solut

ion.c. lowering the temperature of the solution.d. stirring the solution.

1 answer:

joja [24]1 year ago

7 0

To dilute a solute in a solution, it is necessary to add a proper solvent for the reaction to occur, and adding more solvent will cause the solution to dilute even more, therefore the best answer will be letter B

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Is the aqueous solution of each of these salts acidic, basic, or neutral? (a) cr(no3)3 acidic basic neutral (b) nahs acidic basi

melisa1 [442]

1) chromium(III) nitrate is acidic, because it is the salt of weak base (chromium(III) hydroxide Cr(OH)₃) and strong acid (nitric acid HNO₃).
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3) zinc acetate is little basic, because zinc hydroxide (Zn(OH)₂) is stronger base than acetic acid (CH₃COOH).

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3 years ago

Read 2 more answers

How will the calcium atom shown below achieve an octet if it takes part in an ionic bond?

Lelu [443]

Answer:

It will lose 2 electrons

Explanation:

Here we have a calcium atom, which has 20 electrons.

Those 20 electrons are in 4 shells. (2-8-8-2)

Saying there are 2 electrons in the valence shell (Shell N), to get to an octet with an ionic bond (with a group 16 element), those 2 electrons will have to go.

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3 years ago

A box has a volume of 45m3 and is filled with air held at 25∘C and 3.65atm. What will be the pressure (in atmospheres) if the sa

Marina CMI [18]

Answer:

Given:

Initial pressure: $3.65\; \rm atm$ .
Volume was reduced from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ .
Temperature was raised from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ .

New pressure: approximately $3.4\times 10\; \rm atm$ ( $34\; \rm atm$ .) (Assuming that the gas is an ideal gas.)

Explanation:

Both the volume and the temperature of this gas has changed. Consider the two changes in two separate steps:

Reduce the volume of the gas from $45\; \rm m^{3}$ to $5.0\; \rm m^{3}$ . Calculate the new pressure, $P_1$ .
Raise the temperature of the gas from $25\; ^\circ \rm C$ to $35\; ^\circ \rm C$ . Calculate the final pressure, $P_2$ .

By Boyle's Law, the pressure of an ideal gas is inversely proportional to the volume of this gas (assuming constant temperature and that no gas particles escaped or was added.)

For this gas, $V_0 = 45\; \rm m^{3}$ while $V_1 = 5.0\; \rm m^{3}$ .

Let $P_0$ denote the pressure of this gas before the volume change ( $P_0 = 3.65\; \rm atm$ .) Let $P_1$ denote the pressure of this gas after the volume change (but before changing the temperature.) Apply Boyle's Law to find the ratio between $P_1\!$ and $P_0\!$ :

$\displaystyle \frac{P_1}{P_0} = \frac{V_0}{V_1} = \frac{45\; \rm m^{3}}{5.0\; \rm m^{3}} = 9.0$ .

In other words, because the final volume is $(1/9)$ of the initial volume, the final pressure is $9$ times the initial pressure. Therefore:

$\displaystyle P_1 = 9.0\times P_0 = 32.85\; \rm atm$ .

On the other hand, by Amonton's Law, the pressure of an ideal gas is directly proportional to the temperature (in degrees Kelvins) of this gas (assuming constant volume and that no gas particle escaped or was added.)

Convert the unit of the temperature of this gas to degrees Kelvins:

$T_1 = (25 + 273.15)\; \rm K = 298.15\; \rm K$ .

$T_2 = (35 + 273.15)\; \rm K = 308.15\; \rm K$ .

Let $P_1$ denote the pressure of this gas before this temperature change ( $P_1 = 32.85\; \rm atm$ .) Let $P_2$ denote the pressure of this gas after the temperature change. The volume of this gas is kept constant at $V_2 = V_1 = 5.0\; \rm m^{3}$ .

Apply Amonton's Law to find the ratio between $P_2$ and $P_1$ :

$\displaystyle \frac{P_2}{P_1} = \frac{T_2}{T_1} = \frac{308.16\; \rm K}{298.15\; \rm K}$ .

Calculate $P_2$ , the final pressure of this gas:

$\begin{aligned} P_2 &= \frac{308.15\; \rm K}{298.15\; \rm K} \times P_1 \\ &= \frac{308.15\; \rm K}{298.15\; \rm K} \times 32.85\; \rm atm \approx 3.4 \times 10\; \rm atm\end{aligned}$ .

In other words, the pressure of this gas after the volume and the temperature changes would be approximately $3.4\times 10\; \rm atm$ .

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3 years ago

What current (in a) is required to plate out 1. 22 g of nickel from a solution of ni2 in 0. 50 hour?

Mashcka [7]

The current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

<h3>What is current?</h3>

The current is given as the product of the charge with time. In the electrochemical analysis of the nickel, there will be a reduction of the nickel ion to nickel. The formation is given as:

$\rm Ni^{2+}+2e^-\;\rightleftharpoons Ni$

There is the deposition of 1 mole of Ni with 2 electrons transfer. The transfer of charge for 1 mole that is 58.7 grams Nickel is:

$\rm 58.7\;g=2\;\times\;96487\;C\\58.7\;g=192974\;C$

The mass of Ni to be deposited is 1.22 grams. The charge required is given as:

$\rm 58.7\;grams\;Ni=192974\;C\\\\1.22\;grams\;Ni=\dfrac{192974}{58.7}\;\times\;1.22\;C\\\\1.22\;grams\;Ni=4010.7\;C$

The current required to transfer 4010.7 C of charge in 1800 seconds is given as:

$\rm Charge=Current\;\times\;Time\\4010.7\;C=Current\;\times\;1800\;sec\\Current=2.23\;A$

Thus, the current required to accumulate the 1.22 grams of nickel in 0.5 hours is 2.23 A.

Learn more about current, here:

brainly.com/question/23063355

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2 years ago

The advantage of a fixed pulley on a flag pole is that it______.

Finger [1]

Your answer would be A for your homework

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3 years ago